What is the midpoint of the segment shown below ?
1 answer:
Answer:
C) (x,y) = is the mid point of AB.
Step-by-step explanation:
Let A and B be the two end points of the segment shown.
So, the coordinates of the segment AB are A(2,2) and B(3,5).
Let (x,y) be the mid point of AB.
By mid point Formula,
Mid point (x, y) =
⇒(x, y) =
So, x = 5/2, and y = 7/2
Hence, (x,y) = is the mid point of AB.
(c) is the correct option.
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Step-by-step explanation:
This is an interesting question. Wish there were more questions of this kind.
This question helps us recognize the use of the vertex form of a quadratic expression/function.
The vertex form is in the form
The extreme value of the function occurs when x=h, i.e. when the first term vanishes, which leaves the value of the function equal to k. I.e. the vertex of the function is at (h,k).
For example, when
f(x)=5(x-4)^2+7
at x=4, f(4)=5(4-4)^2+7=5(0)+7=7,
in other words, f(x) is at its minimum when x=4, with a value of 7,
even simpler, the vertex of the function is at (4,7).
How do we know if it is a maximum or minimum?
If the first parameter "a" is positive, then at any other value than x=h, the function has a greater value than the vertex, hence a>0 => minimum.
Similarly, if the first parameter "a" is negative, then whenever x does not equal h, the value of the function is smaller, hence a<0 => maximum.
For example, with f(x)=5(x-4)^2+7, a=+5 >0, so (4,7) is a minimum.
Check: f(0)=5(0-4)^2+7=16+7=23 > 7, or (0,23) verifies that (4,7) is a minimum.
For given situations A, B, C, & D, we see that only one function is in the vertex form, i.e. y=a(x-h)+k, namely situation B
B. y=-3(x-2)^2+5, where a=-3, h=2 and k=5.
This means that the function has a maximum at the vertex (2,5). It has a maximum because a=-3 < 0, as discussed above.
Oh yes, enjoy the rest of April Fools Day! lol
This question helps us recognize the use of the vertex form of a quadratic expression/function.
The vertex form is in the form
The extreme value of the function occurs when x=h, i.e. when the first term vanishes, which leaves the value of the function equal to k. I.e. the vertex of the function is at (h,k).
For example, when
f(x)=5(x-4)^2+7
at x=4, f(4)=5(4-4)^2+7=5(0)+7=7,
in other words, f(x) is at its minimum when x=4, with a value of 7,
even simpler, the vertex of the function is at (4,7).
How do we know if it is a maximum or minimum?
If the first parameter "a" is positive, then at any other value than x=h, the function has a greater value than the vertex, hence a>0 => minimum.
Similarly, if the first parameter "a" is negative, then whenever x does not equal h, the value of the function is smaller, hence a<0 => maximum.
For example, with f(x)=5(x-4)^2+7, a=+5 >0, so (4,7) is a minimum.
Check: f(0)=5(0-4)^2+7=16+7=23 > 7, or (0,23) verifies that (4,7) is a minimum.
For given situations A, B, C, & D, we see that only one function is in the vertex form, i.e. y=a(x-h)+k, namely situation B
B. y=-3(x-2)^2+5, where a=-3, h=2 and k=5.
This means that the function has a maximum at the vertex (2,5). It has a maximum because a=-3 < 0, as discussed above.
Oh yes, enjoy the rest of April Fools Day! lol